Changing of index in sum notation

Question

The following is the derivative of the power-series expansion of $e^x$. I can't seem to understand why the starting point changes to $1$ after the second $\text{“}{=}\text{''}.$ Would this surely not be missing out a term? $$ \sum_{n=0}^\infty \frac d {dx} \, \frac{x^n}{n!} = \sum_{n=0}^\infty \frac{nx^{n-1}}{n!} = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{n=0}^\infty \frac{x^n}{n!} = e^x. $$

Answer 1

$\dfrac{nx^{n-1}}{n!} = 0$ when $n=0,$ so the $n=0$ term of the sum $\displaystyle\sum_{n=0}^\infty \frac{nx^{n-1}}{n!}$ can be discarded, yielding $\displaystyle \sum_{n=1}^\infty$ instead of $\displaystyle\sum_{n=0}^\infty.$

Answer 2

Note that $$\sum_{n=0}^{\infty}a_n = a_0 + \sum_{n=1}^{\infty}a_n.$$ Therefore, $$\sum_{n=0}^{\infty}\frac{n x^{n-1}}{n!} = 0+\sum_{n=1}^{\infty}\frac{n x^{n-1}}{n!}.$$

Answer 3

The correct form should have been $$\frac d {dx} e^x =\frac d {dx} \sum_{n=0}^\infty \frac{x^n}{n!}= \sum_{n=0}^\infty \frac d {dx} \, \frac{x^n}{n!} = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{n=0}^\infty \frac{x^n}{n!} = e^x$$

$$\frac {d}{dx} ( 1+x+x^2/2 + x^3/6+\cdots)= 1+x+x^2/2 + x^3/6+\cdots$$